# integral { 1/( (a^2) *cos^2(x) + (b^2)*sin^2(x) ) dx } with a,b out of R\{0}may be that substituting y=tan(X) is of help

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### 1 Answer

Pulling outÂ ` a^2cos^2(x)` we get

`int (dx)/(a^2cos^2(x)+b^2sin^2(x)) = int (dx)/(a^2cos^2(x)(1+b^2/a^2 tan^2(x)))`

This gives us

`1/a^2 int (sec^2(x) dx)/(1+b^2/a^2 tan^2(x))`

Setting `u=b/a tan(x)` , `du = b/a sec^2(x) dx`

or `sec^2(x) dx = a/b du`

We get

`1/a^2 int (a/b du)/(1+u^2)=1/(ab) int (du)/(1+u^2)`

We know this integral is arctan u

So substituting `u = b/a tan(x)` into this equation we get

`int (dx)/(a^2cos^2(x)+b^2cos^2(x)) = 1/(ab) arctan(b/a tan(x)) + C`

You can verify this by taking the derivative.